roberts (eer474) – Quest HW 8 – seckin – (56425)
1
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printout
should
have
16
questions.
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before answering.
001
10.0 points
Find the linearization of
f
(
x
) =
1
√
3 +
x
at
x
= 0.
1.
L
(
x
) =
1
√
3

1
3
x
2.
L
(
x
) =
1
√
3
parenleftBig
1

1
6
x
parenrightBig
correct
3.
L
(
x
) =
1
3
parenleftBig
1

1
3
x
parenrightBig
4.
L
(
x
) =
1
3
parenleftBig
1 +
1
6
x
parenrightBig
5.
L
(
x
) =
1
√
3
parenleftBig
1 +
1
6
x
parenrightBig
6.
L
(
x
) =
1
√
3
+
1
3
x
Explanation:
The linearization of
f
is the function
L
(
x
) =
f
(0) +
f
′
(0)
x .
But for the function
f
(
x
) =
1
√
3 +
x
= (3 +
x
)
−
1
/
2
,
the Chain Rule ensures that
f
′
(
x
) =

1
2
(3 +
x
)
−
3
/
2
.
Consequently,
f
(0) =
1
√
3
,
f
′
(0) =

1
6
√
3
,
and so
L
(
x
) =
1
√
3
parenleftBig
1

1
6
x
parenrightBig
.
002
10.0 points
Use linear appproximation to estimate the
value of 17
1
/
4
. (
Hint
: (16)
1
/
4
= 2.)
1.
17
1
/
4
≈
65
32
correct
2.
17
1
/
4
≈
2
3.
17
1
/
4
≈
31
16
4.
17
1
/
4
≈
33
16
5.
17
1
/
4
≈
63
32
Explanation:
Set
f
(
x
) =
x
1
/
4
, so that
f
(16) = 2 as the
hint indicates. Then
df
dx
=
1
4
x
3
/
4
.
By differentials, therefore, we see that
f
(
a
+ Δ
x
)

f
(
a
)
≈
df
dx
vextendsingle
vextendsingle
vextendsingle
x
=
a
Δ
x
=
Δ
x
4
a
3
/
4
.
Thus, with
a
= 16 and Δ
x
= 1,
17
1
/
4

2 =
1
32
.
Consequently,
17
1
/
4
≈
65
32
.
003
10.0 points
If
f
is the function defined on [

4
,
4] by
f
(
x
) =
x
+

x
 
4
,
which of the following properties does
f
have?
A. Absolute minimum at
x
= 0
.
B. Differentiable at
x
= 0
.
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roberts (eer474) – Quest HW 8 – seckin – (56425)
2
1.
neither of them
2.
both of them
3.
A only
correct
4.
B only
Explanation:
Since

x

=
braceleftbigg
x ,
x
≥
0
,

x ,
x <
0
,
we see that
f
(
x
) =
x
+

x

4 =
braceleftbigg
2
x

4
,
x
≥
0
,

4
,
x <
0
.
Thus on [

4
,
4] the graph of
f
is
2
4

2

4
2
4

2

4
Consequently:
A. True: by inspection.
B. False:
f
is continuous, but not differen
tiable at
x
= 0
.
004
10.0 points
If
f
is the function whose graph is given by
2
4
6
2
4
6
which of the following properties does
f
NOT
have?
1.
differentiable at
x
= 2
correct
2.
local maximum at
x
= 4
3.
f
′
(
x
)
<
0 on (

1
,
2)
4.
lim
x
→
2
+
f
(
x
) =
lim
x
→
2

f
(
x
)
5.
lim
x
→
4
f
(
x
) = 4
Explanation:
The given graph has a removable disconti
nuity at
x
= 4.
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 Spring '09
 Gualdani
 Critical Point, Differential Calculus, Mathematical analysis, abs. max. value

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